Physics question

InlineFour

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Nov 1, 2005
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i'm not sure how to start this. i know that work = joules and work = force x distance.

question: how many joules were consumed?

P = 471 KWH (kilowatt per hour)

471 KWH is consumed in 33 days.

so, 33 days = 792 hours

(471)(792) = joules?
 

PowerMacG5

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Apr 14, 2002
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Originally posted by: InlineFour
i'm not sure how to start this. i know that work = joules and work = force x distance.

question: how many joules were consumed?

P = 471 KWH (kilowatt per hour)

am i missing any other variables?

First of all, kWh = Kilowatt-hour, not kilowatt per hour. kWh is a measurement of energy.

Power (Watts) = Energy (Joule's)/time (seconds).

So, you can rearrange and get Energy (kWh) = power (kW) * time (h).

What you need to know is how much time has passed so you can measure the number of joules.


Also, 1 W = 1 J/s.
 
Dec 10, 2005
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Originally posted by: InlineFour
i'm not sure how to start this. i know that work = joules and work = force x distance.

question: how many joules were consumed?

P = 471 KWH (kilowatt per hour)

471 KWH is consumed in 33 days.

so, 33 days = 792 hours

(471)(792) = joules?

Actually, your answer would have to be in Kilojoules since you are using kilowatts, but beyond that technicallity, it looks fine.
 

PowerMacG5

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Apr 14, 2002
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Originally posted by: InlineFour
the answer in the book is 1.70 x 10^9 J

that's not what i get though.

Thats the answer I get.

First thing I would do is go from kWh to W (J/s). To do this, multiply 471 and 1000, then divide by 33*24. So you get (471*1000)/(33*24)=594.69696969697.

So, 594.69696969697 W = 594.69696969697 J/s. With 33 days times 24 hours, we get 792 hours. Since there are 3600 seconds in an hour, we multiple 792*3600=2,851,200.

So, multiply 594.69696969697 by 2,851,200, and you get 1,695,600,000 = 1.6956 X 10^9 J, or rounding, 1.7 X 10^9 J.

Hooray for me doing your homework.

EDIT: You can do it in less steps than I did, by combining things first, but I took the more straightforward and easier to explain solution. You could have done it without needing to do 2 time calculations.
 

PowerMacG5

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Apr 14, 2002
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Originally posted by: InlineFour
thanks for the thorough explaination. :)

No problem. I am assuming you are taking high school physics. Honestly, stuff like this is not very hard, and if you read your text book you can figure it out.

It also helps knowing what units mean. kWh = kilowatt-hour, if it was kW/h, then it would be kilowatt per hour (which makes no sense, since that would just be some form of Joules).
 

DrPizza

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I think it's been said above, but more simply:

a kilowatt-hour is a measure of energy, not power. A kilowatt-hour is 1000Watts * 3600 seconds.
So, 1 kilowatt-hour = 3,600,000 Joules.
Simply multiply your 471 kilowatt-hours by 3,600,000 Joules per kilowatt-hour.


While powermac may have come up with the correct answer, I, as a physics teacher, would not have awarded full credit for his method (which is quite lacking in units in places.)
For what it's worth, the 33 days has nothing to do with the solution.